DSP Sampling theorem?

Question

Can someone explain this question to me?:

http://img164.imageshack.us/img164/1430/yesxx9.jpg

I don't see how they are equivalent....

Answer 1

Hi!

The trick is to expand y[n], make the substitution Ts = 1/Fs, and then use a little knowledge of periodic signals.

For starters x[n] = A*Cos(2*pi*f0*n*Ts + theta)

now looking at y[n]

y[n] = A*Cos(2*pi*(f0 + L*Fs)*n*Ts + theta)

if we expand the inside brackets we get

y[n] = A*Cos(2*pi*n*Ts*f0 + 2*pi*n*Ts*L*Fs + theta)

we have three terms in side the bracket now, notice that the first and third terms, are the same as the terms in the x[n] functions brackets, only the middle term is new.

We now make the substitution Ts = 1/Fs

y[n] = A*Cos(2*pi*n*Ts*f0 + 2*pi*n*L + theta)

y[n] and x[n] are now the same function except for the "2*pi*n*L"

We must remember that n and L are both integers, so this extra part is is an integer value of 2*pi. seeing as this a cosine function, it repeats every 2*pi. Even though y[n] is now 2*pi out of phase with x[n], they share the same phase and hence can now be considered the same function.